Leetcode19--remove Nth Node from End of List

Given A linked list, remove the nth node from the end of the list and return its head.

For example,

   Given linked list: 1->2->3->4->5, and n = 2.   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:

Given n would always be valid.

Try to do the in one pass.

Main topic

Given a linked list, remove the last nth node and return to the head of the new list note: N is always valid and attempts to scan only one trip.

Difficulty factor:

Easy

Realize

It's easy to just remove the last nth, but if you scan only one trip, I think of a method I'm looking at, and that's the way I look at others.

ListNode *removeNthFromEnd(ListNode *head, int n) {    ListNode *list1, *list2;    list2 = list1 = head;    while (n-- > 0) {        list2 = list2->next;    }    if (list2 == NULL) {        return head->next;    }    while (list2->next != NULL) {        list1 = list1->next;        list2 = list2->next;    }    list1->next = list1->next->next;    return head;}

PS: Deleting a node here should release the memory in it.

Leetcode19--remove Nth Node from End of List